Sum of first n natural numbers

The sum of the first 40 natural numbers is \\(S=\frac{40\times41}{2}=820\\). Half of this sum is \\(\frac{820}{2}=410\\), so option B is correct. Getting 420 would result from an error while finding the sum or taking half of it. Exam tip: remember that the sum of the first n natural numbers is \\(\frac{n(n+1)}{2}\\).

The last (5) terms are (41+42+43+44+45=215). If last terms are asked, finding the total sum is not necessary.

To obtain the sum of the first 21 natural numbers, add the next number, 21, to the sum of the first 20 numbers: \(S_{21}=S_{20}+21=210+21=231\). Hence, 231 is correct. The value 230 would result from adding 20, but the next natural number is 21. Exam tip: For consecutive sums, add the next term to the previous sum.

The sum of the first 36 natural numbers is \,\(S_{36}=\frac{36\times37}{2}=666\), and the sum of the first 30 natural numbers is \,\(S_{30}=\frac{30\times31}{2}=465\). Therefore, the difference is \,\(666-465=201\). It is also the sum of the numbers from 31 to 36. Option 191 would be close to a sum ending at 35, but it does not include 36. Exam tip: The difference of two initial sums equals the sum of the terms left between them.

(S_{39}=\frac{39\times40}{2}=780), so (n=39). In such questions, match (n(n+1)) with (2S_n).

Total dots are (S_{27}=\frac{27\times28}{2}=378). Having (k) dots in the (k)th row shows the sum of natural numbers.

The sum of the first 16 natural numbers is \(S_{16}=\frac{16\times17}{2}=136\). Therefore, \(S_{16}+16=136+16=152\), so option C is correct. The value 136 is only \(S_{16}\); it does not include the additional 16. Exam tip: first use \(S_n=\frac{n(n+1)}{2}\) correctly, then add any extra term separately.

The sum of the first 50 natural numbers is
\(S=\frac{50\times51}{2}=1275\). Therefore, after subtracting 1000, we get \(1275-1000=275\). Hence, option D is correct. A value such as 255 results from an error in finding the sum or carrying out the subtraction. Exam tip: use \(\frac{n(n+1)}{2}\) for the sum of the first \(n\) natural numbers.

The sum of the first k natural numbers is \(S_k=\frac{k(k+1)}{2}\). From \(\frac{m(m+1)}{2}=210\), we get \(m=20\); and from \(\frac{n(n+1)}{2}=325\), we get \(n=25\). Therefore, \(n-m=25-20=5\). Option 4 would result from incorrectly finding the difference between the indices. Exam tip: match a given sum with nearby triangular numbers to identify the index quickly.

The sum is (S_{30}-S_{10}=465-55=410). When it starts from (11), subtract the sum of the first (10) terms.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=44\), we get \(\frac{44\times45}{2}=22\times45=990\). Therefore, 990 is correct. A value such as 968 may result from incorrectly using \(44\times44/2\); the formula must include \(n+1\). Exam tip: When \(n\) is even, divide it by 2 first for quicker calculation.

\(S_{n+1}=1+2+\cdots+n+(n+1)\), whereas \(S_n=1+2+\cdots+n\). Thus, the extra term is \(n+1\). Option B incorrectly uses \(n\) as the added term. Exam tip: identify the newly included term in successive sums.

The sum of the first 60 natural numbers includes the last number, 60. Therefore, subtract 60 to get the sum of the first 59 numbers: \(S_{59}=S_{60}-60=1830-60=1770\). Hence, 1770 is correct. Getting 1780 would mean subtracting 50, which is not the last term here. Exam tip: when reducing a consecutive natural-number sum by one term, subtract the omitted last term.

(S_{14}=105) and (S_{15}=120), so the sum is (225). Find the two similar-looking sums separately.

(S_{42}=\frac{42\times43}{2}=903), so (n=42). Use (2S_n) to see the product of two consecutive numbers.

The marks in the first 15 steps are 1 through 15, so their sum is \(\frac{15\times16}{2}=120\). The next 10 steps contribute zero marks, so they do not change the total. Option 130 would require 10 additional marks, but these steps add zero, not 10. Exam tip: use \(\frac{n(n+1)}{2}\) for the sum of natural numbers from 1 to n.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{52}=\frac{52\times53}{2}=26\times53=1378\). Hence, option B is correct. Option A is a close distractor because it equals the sum up to \(51\): \(\frac{51\times52}{2}=1326\). Exam tip: for the sum of consecutive natural numbers from 1 to \(n\), use \(n(n+1)/2\).

Here, \(S_n\) denotes the sum of the first \(n\) natural numbers. Thus, \(S_{33}=\frac{33\times34}{2}=561\) and \(S_{22}=\frac{22\times23}{2}=253\). Therefore, \(S_{33}-S_{22}=561-253=308\). It is also the sum of the integers from \(23\) to \(33\); an option such as \(298\) can result from an incorrect term count or sum. Exam tip: interpret \(S_m-S_n\) as the sum from \((n+1)\) to \(m\).

\(S_{45}=\frac{45\times46}{2}=1035\), so \(n=45\). Checking options saves time in larger sums.

Here, \(S_n\) denotes the sum of the first \(n\) natural numbers. Adding 20 to \(1+2+3+\cdots+19\) gives \(1+2+3+\cdots+19+20\), which is \(S_{20}\). \(S_{19}\) includes terms only up to 19, so it is not correct. Exam tip: use the last included natural number to identify the subscript of the sum.